CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Thermal Energy

If k 1 = ra...

Question

If $k_{1} =$ rate constant at temperature $T_{1}$ and $k_{2} =$ rate constant at temperature $T_{2}$ for a first-order reaction, then which of the following relations is correct?
[ $E_{a} =$ activation Energy]

A

log \frac{k_{1}}{k_{2}} = \frac{{2.303 E}_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{1} T_{2}}{T_{2} - T_{1}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

log \frac{k_{1}}{k_{2}} = \frac{E_{a}}{2.303 R} (\frac{T_{1} T_{2}}{T_{2} - T_{1}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
Arrhenius gave mathematical expression to deduce the relationship between rate constant and temperature

$k = A e^{- E_{a} R T}$

where,
$A$ is frequency factor and it is a constant
$E_{a}$ is activation energy
$R$ is gas constant, $T$ is temperature

On taking $l n$ on both sides

$ln k = ln A - \frac{E_{a}}{R T}$

If $k_{1}$ and $k_{2}$ are rate constants at temperatures $T_{1}$ and $T_{2}$ respectively, then

$log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{2} T_{1}})$
Hence, the correct option is $B$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Activation energy

(E_{a})

and rate constants (

k_{1}

k_{2}

) of a chemical reaction at two different temperature (

T_{1}

T_{2}

) are related by:

Q. Activation energy

(E_{a})

and rate constant

(k_{1}

k_{2})

of a chemical reaction at two different temperatures

(T_{1}

T_{2})

are related by:

K_{1}

K_{2}

are the equilibrium constants at temperature

T_{1}

T_{2}

T_{2} > T_{1}

Q. The effect of temperature on equilibrium constant is expressed as

(T_{2} > T_{1})

l o g \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

. For endothermic reaction false statement is:

Q. The activation energy for forward and backward reactions are

50 k J / m o l

40 k J / m o l

respectively. If

K_{1}

K_{2}

are the equilibrium constants of reaction at temperature

T_{1}

T_{2}

respectively and

T_{2} > T_{1}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Thermal Energy

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard X Physics

Join BYJU'S Learning Program

CrossIcon